Answer
$V\left( {0,0,c} \right) = - 2\pi \delta RG\ln \left( {\dfrac{{L - c + \sqrt {{R^2} + {{\left( {L - c} \right)}^2}} }}{{ - c + \sqrt {{R^2} + {c^2}} }}} \right)$
Work Step by Step
Recall Eq. (11):
$V\left( P \right) = - G\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \dfrac{{\delta \left( {x,y,z} \right){\rm{d}}S}}{{\sqrt {{{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2} + {{\left( {z - c} \right)}^2}} }}$
Since the surface of the cylinder has a uniform mass distribution, $\delta $ is constant.
Let the axis of the cylinder be the $z$-axis and the point be $P = \left( {0,0,c} \right)$. So, Eq. (11) becomes
$V\left( {0,0,c} \right) = - G\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \dfrac{{\delta {\rm{d}}S}}{{\sqrt {{x^2} + {y^2} + {{\left( {z - c} \right)}^2}} }}$
We express this integral using cylindrical coordinates, where
$x = R\cos \theta $, ${\ \ \ \ \ }$ $y = R\sin \theta $, ${\ \ \ \ \ }$ $z=z$
for $0 \le \theta \le 2\pi $. In cylindrical coordinates, we have $dS = R dz d\theta $. So,
$V\left( {0,0,c} \right) = - G\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{z = 0}^L \dfrac{{\delta R{\rm{d}}z{\rm{d}}\theta }}{{\sqrt {{{\left( {R\cos \theta } \right)}^2} + {{\left( {R\sin \theta } \right)}^2} + {{\left( {z - c} \right)}^2}} }}$
$V\left( {0,0,c} \right) = - G\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{z = 0}^L \dfrac{{\delta R{\rm{d}}z{\rm{d}}\theta }}{{\sqrt {{R^2}{{\cos }^2}\theta + {R^2}{{\sin }^2}\theta + {{\left( {z - c} \right)}^2}} }}$
$V\left( {0,0,c} \right) = - \delta RG\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{z = 0}^L \dfrac{{{\rm{d}}z}}{{\sqrt {{R^2} + {{\left( {z - c} \right)}^2}} }}$
$V\left( {0,0,c} \right) = - 2\pi \delta RG\mathop \smallint \limits_{z = 0}^L \dfrac{{{\rm{d}}z}}{{\sqrt {{R^2} + {{\left( {z - c} \right)}^2}} }}$
Since $\smallint \dfrac{{{\rm{d}}z}}{{\sqrt {{R^2} + {{\left( {z - c} \right)}^2}} }} = \ln \left( {z - c + \sqrt {{R^2} + {{\left( {z - c} \right)}^2}} } \right)$, so
$V\left( {0,0,c} \right) = - 2\pi \delta RG\left[ {\ln \left( {z - c + \sqrt {{R^2} + {{\left( {z - c} \right)}^2}} } \right)} \right]_0^L$
$V\left( {0,0,c} \right) = - 2\pi \delta RG\left[ {\ln \left( {L - c + \sqrt {{R^2} + {{\left( {L - c} \right)}^2}} } \right) - \ln \left( { - c + \sqrt {{R^2} + {c^2}} } \right)} \right]$
$V\left( {0,0,c} \right) = - 2\pi \delta RG\ln \left( {\dfrac{{L - c + \sqrt {{R^2} + {{\left( {L - c} \right)}^2}} }}{{ - c + \sqrt {{R^2} + {c^2}} }}} \right)$
